Bond enthalpies of H_ 2 , X_ 2 and HX are in the ratio 2: 1: 2. If enthalpy of formation of HX is -50 kJmol^ -1 , the bond enthalpy of X_ 2 is

Correct answer: A. 100 kJ mol^ -1 (1)/(2) H_ 2 +(1)/(2) X_ 2 HX Let the bond enthalpy of X-X bond be x . Δ H_ f (HX)=-50=(1)/(2) Δ H_ H-H +(1)/(2) Δ H_ X-X -Δ H_ H-X =(1)/(2) 2 x+(1)/(2) x-2 x=(-x)/(2) x=50 × 2=100 kJ mol^ -1

BITSAT 2012

Exam: BITSAT
Year: 2012
Subject: Chemistry
Chapter: Chemical Bonding and Molecular Structure

BITSAT 2012 · Chemistry · Chemical Bonding and Molecular Structure

Bond enthalpies of H_ 2 , X_ 2 and HX are in the ratio 2: 1: 2. If...

Bond enthalpies of

\mathrm{H}_{2}, \mathrm{X}_{2}

and

\mathrm{HX}

are in the ratio 2: 1: 2. If enthalpy of formation of

\mathrm{HX}

-50 \mathrm{ kJ}\mathrm{mol}^{-1},

the bond enthalpy of

\mathrm{X}_{2}

Options

A $100 \mathrm{ kJ} \mathrm{ mol}^{-1}$
B $300 \mathrm{ kJ} \mathrm{ mol}^{-1}$
C $200 \mathrm{ kJ} \mathrm{ mol}^{-1}$
D $400 \mathrm{ kJ} \mathrm{ mol}^{-1}$

Correct answer

A. $100 \mathrm{ kJ} \mathrm{ mol}^{-1}$

Step-by-step solution

\frac{1}{2} \mathrm{H}_{2}+\frac{1}{2} \mathrm{X}_{2} \longrightarrow \mathrm{HX}

Let the bond enthalpy of

\mathrm{X}-\mathrm{X}

bond be

x

\Delta \mathrm{H}_{\mathrm{f}}(\mathrm{HX})=-50=\frac{1}{2} \Delta \mathrm{H}_{\mathrm{H}-\mathrm{H}}+\frac{1}{2} \Delta \mathrm{H}_{\mathrm{X}-\mathrm{X}}-\Delta \mathrm{H}_{\mathrm{H}-\mathrm{X}}=\frac{1}{2} 2 x+\frac{1}{2} x-2 x=\frac{-x}{2}\therefore x=50 \times 2=100 \mathrm{ kJ} \mathrm{ mol}^{-1}

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